- 博客(4)
- 收藏
- 关注
RSS订阅原创 LeetCode FizzBuzz（C语言）
输出：["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]answer[i] == "FizzBuzz" 如果 i 同时是 3 和 5 的倍数。answer[i] == "Fizz" 如果 i 是 3 的倍数。answer[i] == "Buzz" 如果 i 是 5 的倍数。输出：["1","2","Fizz","4","Buzz"]输出：["1","2","Fizz"]
2023-04-18 16:38:23
308
1
原创 LeetCode FizzBuzz（python）
输出：["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]answer[i] == "FizzBuzz" 如果 i 同时是 3 和 5 的倍数。answer[i] == "Fizz" 如果 i 是 3 的倍数。输出：["1","2","Fizz","4","Buzz"]输出：["1","2","Fizz"]来源：力扣（LeetCode）2023.4.18打卡。
2023-04-18 16:33:41
151
1
原创 LeetCode最富有客户资产总量（C语言）
给你一个 m x n 的整数网格 accounts ，其中 accounts[i][j] 是第 i 位客户在第 j 家银行托管的资产数量。输入：accounts = [[2,8,7],[7,1,3],[1,9,5]]输入：accounts = [[1,5],[7,3],[3,5]]输入：accounts = [[1,2,3],[3,2,1]]第 1 位客户的资产总量 = 1 + 2 + 3 = 6。第 2 位客户的资产总量 = 3 + 2 + 1 = 6。2023.4.18打卡。
2023-04-18 15:55:16
335
1
原创 LeetCode题目一维数组动态和（C语言）
给你一个数组 nums。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i])。请返回 nums 的动态和。
2023-04-18 15:39:08
181
1
空空如也
空空如也
TA创建的收藏夹 TA关注的收藏夹
TA关注的人